Part 2 Continuous functions and their properties

Transcription

1 Part 2 Continuous functions and their properties 2.1 Definition Definition A function f is continuous at a R if, and only if, that is lim f (x) = f (a), x a ε > 0, δ > 0, x, x a < δ f (x) f (a) < ε. Notice that this definition implicitly has three parts. f is defined on some neighbourhood of a, including a, the limit exists, the limit is equal to f (a). Example (i) Prove that the function f : R R given by { 3x if x 0 f (x) = 2x if x > 0 is continuous at x = 0. (ii) Prove, by verifying the ε δ definition, that the function f : R R given by f (x) = x 2 is continuous at a = 2. (iii) Prove that the function f : R R given by { 3x if x 2 f (x) = 0 if x = 2 is not continuous at x = 2. Note From a result in section we have that f is continuous at a if, and only if lim f (x) = f (a) = lim f (x). x a + x a Definition A function f is continuous on an open interval (a, b) if f (x) is defined for, and is continuous at, every x in (a, b), i.e. for all c in (a, b) we have lim x c f (x) = f (c). See Question 1. A function f is continuous on a closed interval [a, b] if f (x) is defined for every x in [a, b], and is continuous at every point in (a, b), i.e. for all 1

2 c in (a, b) we have lim x c f (x) = f (c), while lim x a + f (x) = f (a) and lim x b f (x) = f (b). From a geometric point of view, a function is continuous on an interval if you can draw it s graph without lifting your pencil from the paper. For, if you have to lift your pencil from the graph there has to be jump in the graph. At that point the function is not continuous (the limits of the function as you approach the jump from the right and from the left, if they exist, will be different). To prove a given function is continuous on an interval we need only verify the definition of continuity for each point of the interval. More interesting, perhaps, are examples where continuity fails, and fails frequently. A popular example of such a pathological function is the following, first described by Dirichlet in Example Define f : R R, by { 1 if x is rational f (x) = 0 is x is irrational. Prove that f is nowhere continuous. See Question Continuity Rules We start this section with a lemma almost identical to that in section 1.3. Lemma If f is continuous at c but f (c) 0 then there is a neighbourhood, I say, of c on which f (c) 3 f (c) < f (x) < for all x I. 2 2 In particular, all f (x), x I, have the same sign as f (c). or c I ) ( ( c I ) Proof Not given but virtually identical to that of the Lemma in section 1.3. Given functions continuous at a R we know, in particular, that the limits of the functions exist at a. This means that we can apply the limit rules of section 1.3. In this way we get 2

3 Theorem (i) Suppose that both f and g are continuous at a R. Then Sum Rule: Product Rule: Quotient Rule: f g f + g and f g are continuous at a, fg is continuous at a, is continuous at a, provided g (a) 0. (ii) Suppose that both f and g are continuous on an interval, then f + g, f g, fg and f/g are continuous on the interval with the proviso that in the quotient, g is non-zero on the interval. Proof Not given in this course but virtually identical to that of the similar Theorem in section 1.3. ( ) The observation made in section 1.3 that the results there bore a strong resemblance to results for sequences, from course 153, can also be made here. The similarity comes from Theorem A function f : A R is continuous at x 0 A if, and only if, for all sequences {x n } n 1, x n A if lim n x n = x 0 then lim n f (x n ) = f (x 0 ). Proof Not given in this course. Example Assume the addition formula for sine, namely sin(α + β) = sin α cos β + sin β cos α. ( ) Prove that sin θ is continuous on R. But see Sheet 2 Question 12. Example For the exponential function defined by a power series in course 153, assume that e α+β = e α e β for all real numbers α and β. ( ) A proof of this relies on properties of the Cauchy Product of Series which is not covered in any of our analysis courses. I leave it to the interested student to study this further. ( ) Prove that e x is continuous on R. Example All polynomials are continuous on R. Definition A rational function is a quotient of polynomials. 3

4 ( ) Note: This is obviously similar to the definition of a rational number as the quotient of integers. We found in course 153 that it can take some work to prove a real number is not a rational, just recall the proof that 2 is irrational. Similarly, it can take some work to prove a given function is not rational. For example, could you show that f (x) = x is not a rational function? I leave this to the interested student. But see Question 16 ( ) Theorem A rational function is continuous wherever it is defined. Proof Left to student Theorem Composite Rule for continuous functions. If lim x a g (x) exists and f is continuous at this value, then ( ) lim f (g (x)) = f lim g (x). x a x a Further, if g is continuous at a then lim x a f (g (x)) = f (g (a)), that is, f g is continuous at a. ( ) This is a special case of the Composite Rule for limits. ( ) Example Show that is continuous on R. ( ) 2 + x 2 sin 1 + x Properties of Continuous Functions For the proofs of the next few results we need to recall some definitions and results from course 153. Reminder If S is a set of real numbers then β is called an upper bound for S if x β for all x in S. The least upper bound U for S is an upper bound for S that satisfies U β for all uppers bounds β of S. We write U = sup S or U = lub S. So we can think of the least upper bound as the least of all upper bounds for S. More technically, U = sup S if, and only if, ε > 0, a S : U ε < a U. It is a fundamental property of the real numbers, called Completeness, that if a non-empty set S R has an upper bound then it has a least upper bound. (This was property 10 of R as listed in course 153.) 4

5 Similarly we have lower bound, greatest lower bound, glbs or inf S. And if a non-empty set S R has a lower bound then it has a greatest lower bound. Also, L = inf S if, and only if, ε > 0, a S : L a < L + ε. We say that S is bounded if it has both an upper bound and a lower bound. In the next two results, concerning continuous functions f : [a, b] R, the proofs proceed by first defining sets of the form S = {x : f (x) satisfies some property} [a, b]. This set is obviously bounded above by b. It will be shown that it is a nonempty set, in which case lubs exists. The proofs then concern themselves with this least upper bound. Theorem (Bolzano 1817) Intermediate Value Theorem Suppose that f is a function continuous on a closed interval [a, b] and that f (a) f (b). If γ is some number between f (a) and f (b) then there must be at least one c : a < c < b for which f (c) = γ. In other words, f takes every value between f (a) and f (b). This should reinforce our earlier view that a function is continuous on an interval if, and only if, you can draw it s graph without lifting your pencil from the paper. If you don t lift the pencil from the paper then the graph will cross every horizontal lime from y = f (a) to y = f (b). Example p (x) = x 3 6x x 6. Show that there is a zero of this polynomial between 0 and 4. Is there a zero between 0 and 2.5? There are other interesting corollaries of the intermediate value theorem. Example (i) The image of a continuous function on an interval [a, b] is, itself, an interval. See Question 6. (ii) Fixed Point Theorem. If f : [0, 1] [0, 1] is continuous then there exists c [a, b] such that f (c) = c. See Questions 7 and 18. Definition A function f is said to be bounded on the interval [a, b] if there exist numbers L and U such that L f (x) U for all a x b. Theorem A function continuous on a closed, bounded interval, [a, b], is bounded. Proof Not given in this course but see 5

6 ( ) A proof might proceed by defining S = {x : x [a, b], f bounded on [a, x]}. Obviously a S and S is bounded above by b. Thus lubs exists. One can show (i) If y satisfies a y < lubs then y S. This follows from the definition of lub and S. (ii) Write c = lubs. Then c S. For this you need to show that f is bounded at c. This is achieved by using the left continuity of f at c. (iii) c = b. The method of proof is by contradiction, showing that c [a, b) leads to a contradiction. This is achieved by using the right continuity of f at c. ( ) Definition A function f is said to attain its bounds if there are numbers c and d in the domain such that f (c) f (x) f (d) for all x in the domain. Theorem Boundedness Theorem (1861) A function continuous on a closed, bounded interval [a, b], attains its bounds on that interval. So we can talk about the maximum and minimum values of f. Maximum value = f(d) Minimum value = f(c) c d Note It is important for these last two results that we have f is continuous, the domain, [a, b], is closed, and the domain, [a, b], is bounded. If any of these three conditions fail to hold the conclusion may well not hold. See Question Monotonic functions Definition A function is increasing if f (x 1 ) f (x 2 ) whenever x 1 < x 2, and strictly increasing if f (x 1 ) < f (x 2 ) whenever x 1 < x 2. The obvious definition holds for decreasing and strictly decreasing functions. We say f is (strictly) monotonic if it is either (strictly) increasing or decreasing. 6

7 Example Verify the following. 1. e x is strictly increasing on R, (you may assume again that e x e y = e x+y for all x, y R.) 2. x 3 is strictly increasing on R, See Question x 2 is not increasing on R, See Question for all n N, x n is strictly increasing on [0, ). See Question 11 Theorem Inverse Function Theorem Assume that f is continuous and strictly increasing on [a, b]. Write c = f (a) and d = f (b). Then there exists a function g, continuous and strictly increasing on [c, d] which is inverse to f, i.e. g (f (x)) = x for all x [a, b] and f (g (y)) = y for all y [c, d]. In particular, this result says that a strictly increasing function is a bijection. d y g(y) c f(x) [ ] a x b There is also a version for strictly decreasing functions, and a function for open intervals, where the endpoints might be or. Theorem Inverse Function Theorem for open intervals. Assume that f is continuous and strictly increasing on (a, b) (where possibly a =, b = ). If a is finite write c = f (a) while if a = write c =. If b is finite write d = f (b) while if b = write d =. Then there exists a function g, continuous and strictly increasing on (c, d) which is inverse to f. Proof Not given in this course. There are yet more versions for intervals such as [a, b) or (a, b]. 7

8 Example Each of the functions x 2, x 3,..., x n,... is strictly increasing on [0, ) and continuous there. Therefore the inverse functions x, x 1/3,..., x 1/n,... are well-defined and continuous on [0, ). See Question 13 Example So far we have seen that the function e x is continuous and strictly increasing from (, ) to (0, ). Hence it has an inverse function from (0, ) to (, ). We denote this function by ln x, so ln(e x ) = x for all x (, ) while e ln y = y for all y (0, ). You should ask yourself whether this function, ln, the natural logarithm, has the properties you expect. For example, is it true that ln a + ln b = ln ab for all a, b > 0? Question 14 With the exponential and logarithmic functions we can now define the power of a non-negative number. Definition For a (0, ) and b (, ) define a b = e b ln a. ( ) 2.5 Powers This section will not be covered in the lectures and will not be examined. There is an alternative route to defining a b. For a R and n N we define a n simply to be the product a a... a, n times. Then for a R\ {0}, n N we can define a n = (a n ) 1. For q N and a R, I leave it to the student to show that the equation x q a = 0 has (i) two real roots if q is even and a > 0, (ii) one real root if either a = 0 or q odd and (iii) no real roots if q is even and a < 0. See Question 12 This suggests the following: Definition For q N define a 1/q to be the positive real root of x q a = 0, if q is even and a > 0, the (unique) real root of x q a = 0, if either a = 0 or q odd. So if a < 0 and q is odd then a 1/q is not defined. The next step is when the exponent is a rational number. Definition For r Q, write r = p/q in reduced form with q > 1. Then a r is defined to be ( a 1/q) p. This is meaningful for all a R if q is odd but only for a > 0 if q is even. 8

9 Note: It is important that p/q is in reduced form and the order is as shown, i.e. the 1/q-th root followed by the p-th power. Consider (x 2 ) 1/4. It might be thought that this is the same as x. But they are different. If x < 0 then x 2 > 0 and so (x 2 ) 1/4 is defined, whereas, of course, x fails to exist. What is wrong with (x 2 ) 1/4 is both the order (i.e. square first, fourth root second) and the fact that 2/4 is not in reduced form. Example For all r Q the function f (x) = x r is well-defined and continuous on R if q is odd while if q is even it is well-defined on [0, ) and continuous on (0, ). Question 19 Note You should check that the above definition gives the same value for a r as the earlier definition of a r as e r ln a, when a > 0. Question 20 With the present definition of a r with r Q, it might be asked how to extend the definition to a b with b R, (without using logarithms). For example, what is 2 π? A possible solution to this particular question is that 2 π is the limit of the sequence 2 3, 2 3.1, , , ,.... In each term, the exponent is rational and so the terms can be evaluated by our definition. But we would have to prove that this sequence has a limit. And then we would have to show that if {r n } n 1 is any other sequence of rational numbers satisfying lim n r n = π then lim n 2 rn converges to the same value as the sequence above. Only then would we have a meaningful definition for 2 π. With all this additional work you can see why we took the quick way and defined powers using the exponential and logarithmic functions. ( ) 9

10 Question Sheet 3 1. Write out is symbolic form the definition that f is a continuous function on an open interval (a, b). 2. Sketch the following functions of x R and indicate their discontinuities. (i) [x], the greatest integer less than or equal to x, (ii) x [x] 1 2, (iii) [ x ], x > 0, (iv) [x], x > 0, 1 (v) [ 1 x], 0 < x Prove that h (x) = x is continuous on R. 4. Are the following functions continuous on the domains given or not? Either prove that they are by using the appropriate continuity rules, or show they are not. i) ii) iii) g (x) = 3 + 2x x 2 1 f (x) = x + 2 x on R. firstly on [ 1/2, 1/2], secondly on [ 2, 2]. h (x) = x2 + x 2 (x 2 + 1) (x 1) on R. iv) i (x) = x + 2 x on R. 10

11 v) x + 2 if x < 1 j (x) = x 2 if 1 x 1 x 2 if x > 1. vi) vii) sin x x 0 k (x) = x 1 x = Give examples of l (x) = 1 sin 2x e x on R. a) A strictly increasing function on [0, 1] whose range is not an interval. b) A continuous function on [0, 1] which does not have an inverse. c) A strictly increasing function on ( 1, 1) with range (, ). d) A strictly decreasing function on (0, ) with range ( 1, 0). 6. Prove the Interval Theorem: If f is a continuous function on an interval I then the range, is an interval. f (I) = {f (x) : x I} Here an interval may be finite or infinite, open or closed or half-open. Hint: If f (I) is not bounded below set c = while if f (I) is bounded below set c = glbf (I). If f (I) is not bounded above set d = while if f (I) is bounded above set d = lubf (I). The defining property of an interval J is that if glbj < γ < lubj then γ J. In our case this can be rewritten as: If c < γ < d then γ f (I), that is, γ = f (x) for some x I. You might try to prove this 11

12 by applying the Intermediate Value Theorem to some suitably chosen closed interval. Suitably chosen in that if either c = or d = + the Intermediate Value Theorem cannot be applied directly. 7. Fixed Point Theorem I. If f, g : [a, b] [a, b] are continuous functions such that f (a) g (a) and f (b) g (b) then there exists c [a, b] such that f (c) = g (c). Hint: Consider h (x) = f (x) g (x) and apply the Intermediate Value Theorem. Deduce the Corollary that, if f : [0, 1] [0, 1] is continuous, then there exists c [a, b] such that f (c) = c. 8. Show that 3x = 2 tan x has a solution in the interval (0, π/2). Hint: Be careful. You cannot apply the Fixed Point Theorem I of Question 7 directly, why not? Instead, try to find a subinterval [a, b] (0, π/2) on which the Fixed point Theorem can be applied to give the result. 9. Give examples of i) A function on a closed interval that is not bounded. ii) A continuous function on an open interval that is not bounded. iii) A bounded continuous function on an open interval that does not attain it s bounds. 10. (i) Show that f (x) = 1 x is bounded for all x R. Does it attain its bounds? (ii) Show that f (x) = x x is bounded for all x R. Does it attain its bounds? 12

13 Sketch the graphs of both functions. Hint: Expand and rearrange the inequalities (x 1) 2 0 and (x + 1) Prove that (i) for all n N, with n even, then x n is strictly increasing on [0, ), (ii) for all n N, with n even, then x n is not strictly increasing on R, (iii) for all n N, with n odd, then x n is strictly increasing on R. Hint: use the factorization x n 1 x n 2 = (x 1 x 2 ) ( ) x n x 2 x1 n 2 + x 2 2x n x n 2 2 x 1 + x2 n 1 and show that if x 1 > x 2 then both factors on the right are > 0. In (iii) this is straightforward if x 1 and x 2 are of the same sign, but what can you say if they are of different signs? 12. Prove that for q N and a R, the equation x q a = 0 has (i) exactly two real roots if q is even and a > 0, Hint: First try to apply the Intermediate Value Theorem to f (x) = x q a on [0, ). Unfortunately [0, ) is not a closed interval so something has to be done first. What? (ii) exactly one real root if either a = 0 or q odd, (iii) no real roots if q is even and a < For n N and a R, define a 1/n as the positive real root of x n a = 0, if n is even and a > 0, the (unique) real root of x n a = 0, if either a = 0 or n odd. (i) Prove by verifying the ε δ definition that f (x) = x 1/3 is continuous on R. Hint: Start with x 1/3 a 1/3 = x a x 2/3 + x 1/3 a 1/3 + a 2/3 13

14 Next make sure that x satisfy x a < a /2 which ensures that x and a have the same sign. In that case all of x 2/3, x 1/3 a 1/3 and a 2/3 have the same sign and x 2/3 + x 1/3 a 1/3 + a 2/3 = x 2/3 + x 1/3 a 1/3 + a 2/3 > a 2/3. Thus if a 0. x 1/3 a 1/3 < x a a 2/3 (ii) Prove, by using the results on continuity that if n is odd then x 1/n is continuous on R while if n is even then x 1/n is continuous on (0, ). 14. Prove that the natural logarithm, defined as the inverse of the exponential function, satisfies ln a + ln b = ln ab for all a, b > 0. (As throughout this course you might have to assume that e x e y = e x+y for all x, y R.) 15. (i) For each n N evaluate lim t (ln t) n. t Hint: Look back at Sheet 2 Question 7. (ii) For each n N evaluate lim x (ln x)n. x 0 + Hint: Use (i) and Sheet 1 Question 8. (iii) Evaluate lim xx. x 0 + Hint: Ask yourself how x x, x > 0, is defined. Additional Questions No solutions will be given to these questions, hopefully any hints given will be sufficient. If you need further help, please see me. 14

15 16. Prove that x is not a rational function. Hint: Proof by contradiction. So assume that it is a rational function, that is x2 + 1 = f (x) g (x) for some polynomials f (x) = a n x n +...+a 0 and g (x) = b m x m +...+b 0. Looking at lim x + f (x) /xg (x) will tell us something about m, n, a n and b n. If you then look at lim x f (x) /xg (x) you should be able to get a contradiction. 17. i) Define f : R R, by f (x) = Is this function continuous anywhere? { x if x is rational 1 x if x is irrational. Hint: Look back at Sheet 2 Question 2. So, given an x assume it is continuous there. Take ε = 2x 1 /2 and try to get a contradiction. There is one value of x for which this will not work. ii) Define f : R R, by { 1/q if x Q and x = p/q in reduced form, p 0, f (x) = 0 if x is irrational. Prove that f is continuous at every irrational and discontinuous at every rational. Note. Think of every integer n Z as n 1 Q. In particular 0 = 0 1 and so f (0) = 1 1 = 1. Hint: To prove continuous given any ε > 0 show that there aren t many x with f (x) > ε. To prove discontinuity use proof by contradiction. So if x Q with x = p/q in reduced form, the choice of ε = 1/2q in the definition of continuity will cause problems. 15

16 18. Fixed Point Theorem II. Let f : R R satisfy f (x) f (y) M x y for some 0 M < 1 and all x, y R. We call f a contraction on R. Prove that f has a fixed point. Hint: Assume f (0) > 0, for otherwise look at the function g (x) := f ( x). If there exists a > 0 such that f (a) < a then the result will follow from which Theorem of this course? So assume that f (a) > a for all a > 0. For each n N look at f (n) f (0) = f (n) f (n 1)+f (n 1) f (n 2)+...+f (1) f (0). Use the triangle inequality and the assumptions to deduce n f (0) Mn. Why does this lead to a contradiction? 19. Given r Q \ {0} write r = p/q in reduced form with q > 0. Define a r as ( a 1/q) p, where a 1/q is as defined in 13 and thus meaningfull for all a R when q is odd but only for a 0 when q is even. Prove the following. (i) If q is odd then x r is well-defined and continuous on R. (ii) If q is even then x r (0, ). is well-defined on [0, ) and continuous on 20. For a > 0 and r Q \ {0} we have, above, defined a r. Prove that ln a r = r ln a. Hint: Do this in stages, first for r N and then for r Z. For the next stage, r = 1/n, use the definition of a 1/n as satisfying ( a 1/n) n = a and apply the first step, r N. Combine all steps to get the general case. 16

17 21. (i) Let q N. Justify ( q 1/q 1 ) ( q 1 1/q + q 1 2/q + q 1 3/q ) = q 1. (ii) How many of the terms in are > q 1/2? q 1 1/q + q 1 2/q + q 1 3/q Deduce that ( ) q 1 q 1 1/q + q 1 2/q + q 1 3/q q 1/2. 2 (iii) Conclude that (iv) Evaluate q 1/q 1 2 q 1/2. lim q ( ) 1/q 1. q 17